
Re: Fermat revised ?
Posted:
Aug 15, 2007 11:47 PM


On Aug 15, 8:46 pm, stephane.fr...@gmail.com wrote: > On Aug 15, 6:23 pm, drmwec...@gmail.com wrote: > > > > > > > For starters, yes, there are solutions. > > > The first I found is for n = 2, with x = 27, y = 18, z = 9. > > For n = 3, try x = 256, y = 64, z = 32. > > > Just playing around with QuickBASIC and a calculator program here... > > > Best, Mike > > > Dr. Michael W. Ecker > > Associate Professor of Mathematics > > Pennsylvania State University > > WilkesBarre Campus > > Lehman, PA 18627 > > > On Aug 15, 8:49 pm, stephane.fr...@gmail.com wrote: > > > > I consider the equation : > > > X^n + Y^(n+1) = Z^(n+2) > > > Where X, Y, Z are integer nonnull and n integer > 1 > > > Is there a solution to this equation? > > > > With my pool skill in math, I looked first at X^n + Y^(n+1) = (X^2)^n, > > > chosing Z =X . But I still can see anything > > > > I saw somewhere that if n = 2, then there is solution if X, Y and Z > > > are relatively prime . What about n = 3 ? > > > > SF > > > ??? > > > " I'm not native english speaker so apologize if you can't understand > > > me " Hide quoted text  > > >  Show quoted text  > > I think I understand . > To start a particular case, as there is no constraint on x, y and z , > I can suppose that > x=b*z and y=a*z with a and b integers > Then z^(n+2)=y^(n+1) + x^n <=> z^(n+2) = a^(n+1)*z^(n+1) + b^n * z^n > <=> z^2 = a^(n+1)*z + b^n as z diff 0 > There is solution in real if D=a^2(n+1)+4*b^n >=0, in addition this > need to be a perfect square > The strong condition in this particular case will be to have solution > will be to have a^(n+1)+D^0.5 or a^(n+1)D^0.5 even number > > in your case n=2 , a=2 , b=3 then D=100 then the solutions are 9 or 1 > which provide > 9 => x = 27 , y = 18 , z = 9 > 1 => x = 3, y = 2 , z=3 > > I wonder if there is an easy way to find a, b such as a^(n+1)+D^0.5 > or a^(n+1)D^0.5 is an even number > > SF Hide quoted text  > >  Show quoted text 
Typo error ... 1 => x = 3, y = 2 , z=1 corrected >

